# A triangle has corners at #(9 ,5 )#, #(2 ,3 )#, and #(7 ,6 )#. What is the area of the triangle's circumscribed circle?

##### 2 Answers

#### Explanation:

The standard Cartesian form for the equation of a circle is:

where

We can use the points,

Expand the squares:

Subtract equation [3.1] from equation [2.1]:

Combine like terms:

Subtract equation [3.1] from equation [4.1]:

Combine like terms:

Multiply both sides of equation [5] by

#72 - 10h -6k - 3/2(93 - 14h - 4k) = 0

Distribute the

#72 - 10h -6k - 279/2 + 21h + 6k = 0

Combine like terms:

Use equation [6] to find the value of k:

Use equation [2] to find the value of

Because the area of a circle is

Circum center **(6.1364, 1.7727)**

Area of circumcircle **58.5065**

#### Explanation:

Slope of AB m1 = (3-5)/(2-9)=2/7.

Slope of perpendicular at mid point of AB = -1/m1 = -7/2

Midpoint of AB = (9+2)/2, (3+5)/2 = 11/2, 4

Eqn of perpendicular bisector of AB is

Slope of BC m2 = (6-3)/(7-2) = 3/5.

Slope of perpendicular at mid point of AB = -1/m1 = -5/3.

Midpoint of BC = (7+2)/2, (6+3)/2 = 9/2, 9/2

Eqn of perpendicular bisector of BC is

Solving Eqns (1), (2)

Circum center **(6.1364, 1.7727)**

Area of circumcircle